The osmotic pressures of 0.010 M solutions of KI and sucrose (`C_(12)H_(22)O_(11)`) are 0.432 atm and 0.24 atm respectively. The van't Hoff fector for KI is :

Options

A. 1.8
B. 0.8
C. 1.2
D. 1

Answer

To find the van't Hoff factor (i) for KI based on the given osmotic pressures, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Osmotic Pressure**:
The osmotic pressure (π) is given by the formula:
\[ \pi = C \cdot R \cdot T \cdot i \]
where:
- \( \pi \) = osmotic pressure
- \( C \) = concentration of the solution (in molarity)
- \( R \) = universal gas constant
- \( T \) = temperature (in Kelvin)
- \( i \) = van't Hoff factor

2. **Set Up the Equations for KI and Sucrose**:
For the 0.010 M solution of KI:
\[ \pi_{KI} = 0.010 \cdot R \cdot T \cdot i_{KI} \]
For the 0.010 M solution of sucrose (which does not dissociate):
\[ \pi_{sucrose} = 0.010 \cdot R \cdot T \cdot i_{sucrose} \]
Since sucrose is a non-electrolyte, its van't Hoff factor \( i_{sucrose} = 1 \):
\[ \pi_{sucrose} = 0.010 \cdot R \cdot T \cdot 1 \]

3. **Substitute the Given Values**:
The osmotic pressures are given as:
- \( \pi_{KI} = 0.432 \, \text{atm} \)
- \( \pi_{sucrose} = 0.24 \, \text{atm} \)

4. **Write the Equations**:
From the equations:
\[ \pi_{KI} = 0.010 \cdot R \cdot T \cdot i_{KI} \]
\[ \pi_{sucrose} = 0.010 \cdot R \cdot T \]

5. **Divide the Two Equations**:
To find \( i_{KI} \), we can divide the two equations:
\[ \frac{\pi_{KI}}{\pi_{sucrose}} = \frac{0.010 \cdot R \cdot T \cdot i_{KI}}{0.010 \cdot R \cdot T} \]
This simplifies to:
\[ \frac{\pi_{KI}}{\pi_{sucrose}} = i_{KI} \]

6. **Calculate the Van't Hoff Factor**:
Substitute the values of osmotic pressures:
\[ i_{KI} = \frac{0.432}{0.24} \]
Calculate the ratio:
\[ i_{KI} = 1.8 \]

### Final Answer:
The van't Hoff factor for KI is \( i_{KI} = 1.8 \).

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